c) The total resistance for all contributors in series is just the sum:

4 ohms + 3 ohms + 1 ohm = 8 ohms

2) The photo shows a sketch of a Wheatstone Bridge Circuit at Harrison College, to be used to find the resistances of two resistors (R1 and R2) connected as shown, with two differing positions of the galvanometer. The total resistance R is also to be taken.

A student using the diagram finds his voltmeter reads 1.0 V and the total resistance is 4 ohms when the slide wire is at the end position noted.

If the balance for obtaining G = 0 (e.g. galvanometer reading zero) is the intermediate position, and the slide wire is then at 45 cm, find the values of R1 and R2.

The total resistance from the end position (100 cm) = 4 ohms. We know from the photo that the two resistors R1 and R2 are connected in series so that:

R1 + R2 = 4 ohms

We are given the first position of the slide wire as L1 = 45 cm, then:

R1/ 4 ohms = 45 cm/ 100 cm

and: R1 = 0.45 (4 ohms) = 1.8 ohms

Then:

R2 = 4 ohms - 1.8 ohms = 2.2 ohms

3) A resistance R2 is connected in parallel with a resistance R1. What resistance R3 must be connected in series with the combination of R1 and R2 so that the equivalent resistance is equal to the resistance R1? Draw a circuit diagram of the arrangement.

We let R1 = r and let R2 = 2r.

These are in parallel so the total resistance for them is:

R(T) = R1 R2/ (R1 + R2) = r(2r)/ (r + 2r) = 2r^2/ 3r = 2r/3

We require that the total resistance in series be such that the equivalent resistance is equal to the resistance R1, or r.

Thus, we require:

R3 + 2r/3 = r

and, solving by algebra:

R3 = r - 2r/3 = r/3

The circuit diagram is shown, labelled as 'Prob. 3'.

4) Three equal resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 10 watts. (Note: Power = V x I, voltage x current).

What power would be dissipated if the three resistors were connected in parallel across the same potential difference?

You need to bear in mind that for the case in parallel the current is divvied up, e.g. total current I = I1 + I2 + I3. Since all the resistors are equal (call each r ohms) and the voltage is the same then the power for the parallel case would be one third the power for the series case, or 10/3 Watts.

About Me

Specialized in space physics and solar physics, developed first astronomy curriculum for Caribbean secondary schools, has written thirteen books - the most recent:Fundamentals of Solar Physics. Also: Modern Physics: Notes, Problems and Solutions;:'Beyond Atheism, Beyond God', Astronomy & Astrophysics: Notes, Problems and Solutions', 'Physics Notes for Advanced Level&#39, Mathematical Excursions in Brane Space, Selected Analyses in Solar Flare Plasma Dynamics; and 'A History of Caribbean Secondary School Astronomy'. It details the background to my development and implementation of the first ever astronomy curriculum for secondary schools in the Caribbean.